Schrijver
| The "I am very bored" thread!
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[D-Tail]
 msx guru Berichten: 3017 | Geplaatst: 07 Juni 2006, 00:00   |
You're totally wrong here, but it's fun guessing, right? I'll post the right answer in 24 hours, if not found out before  |
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wolf_
 msx legend Berichten: 4713 | Geplaatst: 07 Juni 2006, 00:42   |
o_O A planet in a universe..
Can one get there with a transdimensional engine, containing bugs?  |
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someone2 msx user Berichten: 33 | Geplaatst: 07 Juni 2006, 00:51   |
Hmm... so, with "wolf_ wouldn't exist there" you actually mean wolf_ the MSXer, not wolves?  This is getting tricky  |
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Edwin msx professional Berichten: 620 | Geplaatst: 07 Juni 2006, 01:02   |
I would say an "F".
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[D-Tail]
 msx guru Berichten: 3017 | Geplaatst: 07 Juni 2006, 09:12   |
Edwin, you got it right! Congrats man!  |
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AuroraMSX
 msx master Berichten: 1248 | Geplaatst: 07 Juni 2006, 09:19   |
Quote:
| This one is specifically for AuroraMSX, (because brute force will never work here  )
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Tsk, as long as you can write a nice math equation for it, it can be solved with brute force:
Iterate over N: (N*10 + 7) * 7 == int(log10(N)+1)*7 + N
I'll leave the programming as an exercise for the reader  |
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norakomi msx professional Berichten: 861 | Geplaatst: 07 Juni 2006, 10:49   |
Quote:
| Nooooooooooooo... a number cannot start with zero  You are on the right track though  But, it is not that simple
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Doh
Well if I would continue this sequence I will find the number, or is my theory not correct.
Then it might get interesting after all !! haha
ok, continuing sequence:
00.000*7=0
0+40.579=40.579
400.000*7=2.800.000
2.800.000+40.579=2.840.579
8.000.000*7=56.000.000
56.000.000+2.840.579=58.840.579
80.000.000*7=560.000.000
560.000.000+58.840.579=618.840.579
100.000.000*7=700.000.000
700.000.000+618.840.579=1.318.840.579
3.000.000.000*7=21.000.000.000
21.000.000.000+1.318.840.579=22.318.840.579
20.000.000.000*7=140.000.000.000
140.000.000.000+22.318.840.579=162.318.840.579
600.000.000.000*7=4.200.000.000.000
4.200.000.000.000+162.318.840.579=4.362.318.840.579
pffff, now it gets boring.
I think i did something wrong
one hint please:
tell me,
does you number end with ...... 3.623.188.405.797 ???
(because multiply this with 7 and you get 25. 362.318.840.579)
and... your number starts with 11 , right ? |
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norakomi msx professional Berichten: 861 | Geplaatst: 07 Juni 2006, 10:54   |
Quote:
| Now, about the sequence, that was really tough. I had to plot them on Matlab, and they looked like a quadratic function, and then out of nowhere, I saw it
0 = 0 * 1 * 2
6 = 1 * 2 * 3
24 = 2 * 3 * 4
60 = 3 * 4 * 5
120 = 4 * 5 * 6
210 = 5 * 6 * 7
which means the next number should be
6 * 7 * 8 = 336 
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YEAAAAAAAAAH !!! goooood job !! |
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someone2 msx user Berichten: 33 | Geplaatst: 07 Juni 2006, 15:36   |
Quote:
| one hint please:
tell me,
does you number end with ......3.623.188.405.797 ???
(because multiply this with 7 and you get 25.362.318.840.579)
and... your number starts with 11 , right ?
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Yes, you are absolutely right. These are the last 13 digits  You are getting close too, it doesn't take much longer from now
But, no the number doesn't start with 11! Sorry  |
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someone2 msx user Berichten: 33 | Geplaatst: 07 Juni 2006, 15:38   |
Quote:
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Tsk, as long as you can write a nice math equation for it, it can be solved with brute force:
Iterate over N: (N*10 + 7) * 7 == int(log10(N)+1)*7 + N
I'll leave the programming as an exercise for the reader 
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No. as you can see the number is larger than 13 digits. It can't even be represented as a long long!  So, iterating is going to take billions of years  |
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AuroraMSX
 msx master Berichten: 1248 | Geplaatst: 07 Juni 2006, 16:26   |
Quote:
| Quote:
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Tsk, as long as you can write a nice math equation for it, it can be solved with brute force:
Iterate over N: (N*10 + 7) * 7 == int(log10(N)+1)*7 + N
I'll leave the programming as an exercise for the reader 
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No. as you can see the number is larger than 13 digits. It can't even be represented as a long long!  So, iterating is going to take billions of years 
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So what? You stated the problem cannot be solved with brute force, I just showed that it is possible -- you just need a bit of patience
(Or you could convert norakomi's pattern searching into an algorithm... I not so bored that I'm going to post that, but it is possible and will find the number faster than brute force) |
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AuroraMSX
 msx master Berichten: 1248 | Geplaatst: 07 Juni 2006, 16:53   |
Here's another one. How does this sequence continue?
0001001101011110...
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someone2 msx user Berichten: 33 | Geplaatst: 07 Juni 2006, 16:56   |
Quote:
| Here's another one. How does this sequence continue?
0001001101011110...
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Hmm...
0001001101011110
could be decomposed to
0001 0011 0101 1110
1 3 5 7
so, it should continue
0001001101011110 1001 1011... |
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someone2 msx user Berichten: 33 | Geplaatst: 07 Juni 2006, 16:58   |
And speaking of sequences, here is another one
4, 5, 9, 11, 12, 13, 14.....
How should it continue?
(Hint: This is called the "Even Steven" sequence!)
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AuroraMSX
 msx master Berichten: 1248 | Geplaatst: 07 Juni 2006, 18:19   |
Quote:
| I not so bored that I'm going to post that, but it is possible and will find the number faster than brute force)
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Ah well, what the heck. Here goes - in Java for a change
public class Norakomis7{
public static void main(String[] args) {
int lastDigit = 7;
int carry = 0;
int value;
boolean ready;
Vector<Integer> digits = new Vector<Integer>();
digits.add(new Integer(lastDigit));
do {
value = lastDigit * 7 + carry;
lastDigit = value % 10;
carry = value/10;
ready = (lastDigit == 7 && carry == 0);
if(!ready)
digits.add(0, new Integer(lastDigit));
} while(!ready);
for(Iterator<Integer> it = digits.iterator(); it.hasNext(); )
{
System.out.print(it.next());
}
System.out.println();
}
}
The output is 1014492753623188405797
As for the sequence, your solution sounds valid, but is not what I anticipated:
0001 0011 0101 1110 0010 0110 1011 1100 ...
Now, how does the sequence continue?
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